Câu `1`:
`(x-3)(x+1)(8-2x)=0`
⇔`2(x-3)(x+1)(4-x)=0`
⇔\(\left[ \begin{array}{l}x-3=0\\x+1=0\\4-x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-1\\x=4\end{array} \right.\)
Vậy `S={3,-1,4}`
Câu `2`:
`(x-1)(x+5=0`
⇔\(\left[ \begin{array}{l}x-1=0\\x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.\)
Vậy `S={1,-5}`
Câu `3`:
`x^2-2x+1=0`
⇔`(x-1)^2=0`
⇔`x-1=0`
⇔`x=1`
Vậy `S={1}`